An element $e\in R$ is called idempotent if $e^2=e$. Assume that $e$ is idempotent and $er=re$ $\forall r\in R$. Prove that $Re$ and $R(1-e)$ are two sided ideals of $R$ and $R\cong Re\times R(1-e)$.
I have successfully proved the other results and stuck on the last one. Applying Chinese Remainder Theorem we have that $R/(Re\cap R(1-e))\cong R/Re\times R/R(1-e)$ .
Also $Re+R(1-e)=R$ and $Re\cap R(1-e)=\{0\}$. Thus $R\cong R/Re\times R/R(1-e)$.
where am I going wrong?
Best Answer
Let $f=1-e$, so that $ef=fe$ and $f^2=f,e^2=e$. We have a morphism of rings (prove it is one!) $R\to Re\times Rf$ that sends $a=ae+af$ to $ (af,ae)$. The rings $Re$ and $Rf$ have units $e,f$ respectively. This is an epimorphism since $ae+bf$ maps to $(ae,bf)$ (for $ef=fe$ and $f,e$ are idempotent), and a monomorphism since $af=ae=0$ gives $af+ae=a=0$. The point of this is you don't need to meddle with CRT and all those isomorphism theorems: central idempotents disconnect your ring "nicely" and you can make concrete computations with them.